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2.5 Inverse Matrices 2.5. Inverse Matrices 81 2.5 Inverse Matrices Suppose A is a square matrix. We look for an “inverse matrix” A1 of the same size, such that A1 times A equals I . Whatever A does, A1 undoes. Their product is the identity matrix—which does nothing to a vector, so A1Ax D x. But A1 might not exist. What a matrix mostly does is to multiply a vector x. Multiplying Ax D b by A 1 gives A1Ax D A1b. This is x D A1b. The product A1A is like multiplying by a number and then dividing by that number. A number has an inverse if it is not zero— matrices are more complicated and more interesting. The matrix A1 is called “A inverse.” DEFINITION The matrix A is invertible if there exists a matrix A1 such that A1A D I and AA1 D I: (1) Not all matrices have inverses. This is the first question we ask about a square matrix: Is A invertible? We don’t mean that we immediately calculate A1. In most problems we never compute it! Here are six “notes” about A1. Note 1 The inverse exists if and only if elimination produces n pivots (row exchanges are allowed). Elimination solves Ax D b without explicitly using the matrix A 1. Note 2 The matrix A cannot have two different inverses. Suppose BA D I and also AC D I . Then B D C , according to this “proof by parentheses”: B.AC/ D .BA/C gives BI D IC or B D C: (2) This shows that a left-inverse B (multiplying from the left) and a right-inverse C (multi- plying A from the right to give AC D I ) must be the same matrix. Note 3 If A is invertible, the one and only solution to Ax D b is x D A1b: Multiply Ax D b by A1: Then x D A1Ax D A1b: Note 4 (Important) Suppose there is a nonzero vector x such that Ax D 0. Then A cannot have an inverse. No matrix can bring 0 back to x. If A is invertible, then Ax D 0 can only have the zero solution x D A10 D 0. Note 5 A 2 by 2 matrix is invertible if and only if ad bc is not zero: Ä Ä ab1 1 d b 2 2 D : by Inverse: cd ad bc ca (3) This number ad bc is the determinant of A. A matrix is invertible if its determinant is not zero (Chapter 5). The test for n pivots is usually decided before the determinant appears. 82 Chapter 2. Solving Linear Equations Note 6 A diagonal matrix has an inverse provided no diagonal entries are zero: 2 3 2 3 d1 1=d1 6 : 7 1 6 : 7 If A D 4 :: 5 then A D 4 :: 5 : dn 1=dn A D 12 Example 1 The 2 by 2 matrix 12 is not invertible. It fails the test in Note 5, because ad bc equals 2 2 D 0. It fails the test in Note 3, because Ax D 0 when x D .2; 1/. It fails to have two pivots as required by Note 1. Elimination turns the second row of this matrix A into a zero row. The Inverse of a Product AB For two nonzero numbers a and b, the sum a C b might or might not be invertible. The a D 3 b D3 1 1 a C b D 0 numbers and have inverses 3 and 3 . Their sum has no inverse. ab D9 1 1 But the product does have an inverse, which is 3 times 3 . For two matrices A and B, the situation is similar. It is hard to say much about the invertibility of A C B. But the product AB has an inverse, if and only if the two factors A and B are separately invertible (and the same size). The important point is that A1 and B1 come in reverse order: If A and B are invertible then so is AB. The inverse of a product AB is .AB/1 D B1A1: (4) To see why the order is reversed, multiply AB times B 1A1. Inside that is BB 1 D I : Inverse of AB .AB/.B1A1/ D AIA1 D AA1 D I: We moved parentheses to multiply BB 1 first. Similarly B 1A1 times AB equals I . This illustrates a basic rule of mathematics: Inverses come in reverse order. It is also common sense: If you put on socks and then shoes, the first to be taken off are the . The same reverse order applies to three or more matrices: Reverse order .ABC /1 D C 1B1A1: (5) Example 2 Inverse of an elimination matrix.IfE subtracts 5 times row 1 from row 2, then E1 adds 5 times row 1 to row 2: 2 3 2 3 100 100 E D 45105 and E1 D 45105 : 001 001 Multiply EE1 to get the identity matrix I . Also multiply E 1E to get I . We are adding and subtracting the same 5 times row 1. Whether we add and then subtract (this is EE 1/ or subtract and then add (this is E 1E/, we are back at the start. 2.5. Inverse Matrices 83 For square matrices, an inverse on one side is automatically an inverse on the other side. If AB D I then automatically BA D I . In that case B is A1. This is very useful to know but we are not ready to prove it. Example 3 Suppose F subtracts 4 times row 2 from row 3, and F 1 adds it back: 2 3 2 3 100 100 F D 40105 and F 1 D 40105 : 0 41 041 Now multiply F by the matrix E in Example 2 to find FE. Also multiply E 1 times F 1 to find .FE/1. Notice the orders FE and E1F 1! 2 3 2 3 10 0 100 FE D 4 51 05 is inverted by E1F 1 D 4 5 105 : (6) 20 41 0 4 1 The result is beautiful and correct. The product FE contains “20” but its inverse doesn’t. E subtracts 5 times row 1 from row 2. Then F subtracts 4 times the new row 2 (changed by row 1) from row 3. In this order FE, row 3 feels an effect from row 1. In the order E1F 1, that effect does not happen. First F 1 adds 4 times row 2 to row 3. After that, E1 adds 5 times row 1 to row 2. There is no 20, because row 3 doesn’t change again. In this order E 1F 1,row3 feels no effect from row 1. In elimination order F follows E. In reverse order E1 follows F 1. E 1F 1 is quick. The multipliers 5, 4 fall into place below the diagonal of 1’s. This special multiplication E 1F 1 and E1F 1G1 will be useful in the next sec- tion. We will explain it again, more completely. In this section our job is A1, and we expect some serious work to compute it. Here is a way to organize that computation. Calculating A1 by Gauss-Jordan Elimination I hinted that A1 might not be explicitly needed. The equation Ax D b is solved by x D A1b. But it is not necessary or efficient to compute A1 and multiply it times b. Elimination goes directly to x. Elimination is also the way to calculate A 1,aswenow show. The Gauss-Jordan idea is to solve AA1 D I , finding each column of A1. 1 A multiplies the first column of A (call that x1/ to give the first column of I (call that e1/. This is our equation Ax1 D e1 D .1;0;0/. There will be two more equations. 1 Each of the columns x1, x2, x3 of A is multiplied by A to produce a column of I : 1 1 3 columns of A AA D A x1 x2 x3 D e1 e2 e3 D I: (7) To invert a 3 by 3 matrix A, we have to solve three systems of equations: Ax 1 D e1 and 1 Ax2 D e2 D .0;1;0/and Ax3 D e3 D .0;0;1/. Gauss-Jordan finds A this way. 84 Chapter 2. Solving Linear Equations The Gauss-Jordan method computes A1 by solving all n equations together. Usually the “augmented matrix” ŒA b has one extra column b. Now we have three right sides e1; e2; e3 (when A is 3 by 3). They are the columns of I , so the augmented matrix is really the block matrix ŒA I . I take this chance to invert my favorite matrix K, with 2’s on the main diagonal and 1’s next to the 2’s: 2 3 2 1 01 0 0 K 6 7 Start Gauss-Jordan on K e1 e2 e3 D 4 1 2 10 1 05 0 1 20 0 1 2 3 2 10100 ! 4 3 1 5 . 1 C / 0 2 1 2 10 2 row 1 row 2 0 12001 2 3 2 10100 4 0 3 1 1 105 ! 2 2 4 1 2 . 2 C / 003 3 3 1 3 row 2 row 3 We are halfway to K1. The matrix in the first three columns is U (upper triangular). The 2; 3 ; 4 pivots 2 3 are on its diagonal.
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